Wednesday, July 05, 2017

I've been trying to relearn some of the combinatorics I used to know, I redid a nice recursive algorithm (from early university days), with python it looks so beautiful and nice. A simple implementation is below, so nice and simple.

The output lends itself to a simple and nice iterative algorithm, I'll follow up on.

The interesting follow up to combinatorial generation algorithms is using ranking and unranking. At this point, I think we need a O(n) to identify if two combinations are the same -- given in any order of elements. For example to compare
$\{1, 2, 3\}, \{1, 3, 2\}, \{2, 1, 3\}, \{3, 1, 2\}, \{2, 3, 1\}, \{3, 2, 1\}$
we need to identify them and associate the same rank with them. I think for a subset of size $k$, the cost of ranking is $\theta(k)$, but I need to see if there is a better way to do it.